Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<set>

#define fst first
#define snd second

using namespace std;

typedef pair<int,int> pp;
vector<int> G[105000];
int dis[105000];
int S[105000];

int main(void) {
    int n,m,s,t;
    scanf("%d%d%d%d",&n,&m,&s,&t);
    s--, t--;
    memset(dis, 0x20, sizeof(dis));
    for(int i=0; i<m; i++) {
        int x,y;
        scanf("%d%d",&x,&y);
        x--, y--;
        G[x].push_back(y);
        G[y].push_back(x);
    }

    {
        priority_queue<pp> q;
        q.push(pp(-0, s));

        while(!q.empty()) {
            pp t = q.top(); q.pop();
            int d = -t.fst, x = t.snd;

            if(dis[x] != 0x20202020) continue;
            dis[x] = d;
            S[d]++;

            for(int y : G[x])
                if(dis[y] == 0x20202020)
                    q.push(pp(-(d+1), y));
        }
    }

    int old = dis[t];
    memset(dis, 0x20, sizeof(dis));

    long long int res=0;
    {
        priority_queue<pp> q;
        q.push(pp(-0, t));

        while(!q.empty()) {
            pp t = q.top(); q.pop();
            int d = -t.fst, x = t.snd;

            if(dis[x] != 0x20202020) continue;
            dis[x] = d;

            if(old-2-d >= 0)
                res += S[old-2-d];

            for(int y : G[x])
                if(dis[y] == 0x20202020)
                    q.push(pp(-(d+1), y));
        }
    }

    printf("%lld\n", res);
    return 0;

}